$f(x) = |3x+3|$ Evaluate the definite integral. $\int^1_{-3}f(x)\,dx = $ Choose 1 answer: Choose 1 answer: (Choice A) A $7$ (Choice B) B $12$ (Choice C) C $18$ (Choice D) D $30$
Solution: Splitting up the absolute value Notice that the absolute value function is a piecewise function. Here we have that: $f(x) = \begin{cases} 3x+3 & \text{for} ~~~~x\geq-1 \\ -3-3x & \text{for} ~~~~ x \lt-1\end{cases}$ Splitting up the definite integral Since we're working with a piecewise function, we need to split the definite integral up into two pieces: $\phantom{=} \int^1_{-3}f(x)\,dx$ $= \int^1_{-1}f(x)\,dx + \int^{-1}_{-3}f(x)\,dx~~~~~~$ [Why did we split at -1?] $= \int^1_{-1}(3x + 3)\,dx + \int^{-1}_{-3}(-3-3x)\,dx ~~~~~~$ Evaluating each piece Next, let's evaluate each of these definite integrals one at a time. The first definite integral: $\begin{aligned} \int^1_{-1}(3x + 3)\,dx~ &=\dfrac32x^2 + 3x\Bigg|^1_{{-1}} \\\\ &= \left[\dfrac32 \cdot( 1)^2 + 3\cdot(1) \right] - \left[\dfrac32\cdot({-1})^2 + 3\cdot({-1}) \right] \\\\ &= \left[\dfrac{9}{2}\right] -\left[-\dfrac32 \right] \\\\ &= {6}\end{aligned}$ The second definite integral: $\begin{aligned} \int^{-1}_{-3}(-3-3x)\,dx~ &=-3x - \dfrac32x^2\Bigg|^{-1}_{{-3}} \\\\ &= \left[- 3\cdot({-1})-\dfrac32 \cdot( {-1})^2 \right] - \left[- 3\cdot({-3})-\dfrac32\cdot({-3})^2 \right] \\\\ &= \left[\dfrac{3}{2}\right] -\left[-\dfrac{9}2 \right] \\\\ &= {6}\end{aligned}$ Putting the pieces together Now let's add these two pieces together to find the answer: $\phantom{=} \int^1_{-1}(3x + 3)\,dx + \int^{-1}_{-3}(-3-3x)\,dx$ $ = {6} + {6}$ $ = 12$ The answer $\int^1_{-3}f(x)\,dx = 12$